Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int tinf = (int)1e9 + 7;
const long long inf = (long long)1e18 + 7;
const int N = 2e5 + 5;
const long long M = 1e9 + 7;
const int rx[4] = {1, 0, -1, 0};
const int ry[4] = {0, 1, 0, -1};
int tin[N], tout[N], timer, up[N][20], n, m;
int num[N];
vector<int> g[N];
vector<int> p[N];
vector<int> ans;
bool used[N];
struct node {
  int sum;
  int l, r;
  node() : l(0), r(0), sum(0) {}
};
vector<node> all;
vector<int> root;
int add_node() {
  all.push_back(node());
  return (int)all.size() - 1;
}
int add_node(int v) {
  all.push_back(all[v]);
  return (int)all.size() - 1;
}
void build(int v, int tl, int tr) {
  if (tl == tr) return;
  int p = add_node();
  all[v].l = p;
  p = add_node();
  all[v].r = p;
  int tm = (tl + tr) >> 1;
  build(all[v].l, tl, tm);
  build(all[v].r, tm + 1, tr);
}
int update_per(int v, int tl, int tr, int p, int x) {
  v = add_node(v);
  if (tl == tr) {
    all[v].sum += x;
    return v;
  }
  int tm = (tl + tr) >> 1;
  if (p <= tm) {
    int cur = update_per(all[v].l, tl, tm, p, x);
    all[v].l = cur;
  } else {
    int cur = update_per(all[v].r, tm + 1, tr, p, x);
    all[v].r = cur;
  }
  all[v].sum = all[all[v].r].sum + all[all[v].l].sum;
  return v;
}
bool upper(int v, int u) { return tin[v] <= tin[u] && tout[v] >= tout[u]; }
int lca(int v, int u) {
  if (upper(v, u)) return v;
  if (upper(u, v)) return u;
  for (int i = 19; i >= 0; i--) {
    if (!upper(up[v][i], u)) {
      v = up[v][i];
    }
  }
  return up[v][0];
}
void dfs(int v, int par = 0) {
  used[v] = 1;
  tin[v] = ++timer;
  up[v][0] = par;
  for (int i = 1; i < 20; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
  }
  if ((int)p[v].size()) {
    for (int i = 0; i < (int)p[v].size(); i++) {
      root.push_back(
          update_per(root[(int)root.size() - 1], 0, m - 1, p[v][i], 1));
    }
  } else
    root.push_back(update_per(root[(int)root.size() - 1], 0, m - 1, 0, 0));
  num[v] = root.back();
  for (auto j : g[v]) {
    if (!used[j]) {
      dfs(j, v);
    }
  }
  tout[v] = ++timer;
  if ((int)p[v].size()) {
    for (int i = 0; i < (int)p[v].size(); i++) {
      root.push_back(
          update_per(root[(int)root.size() - 1], 0, m - 1, p[v][i], -1));
    }
  } else
    root.push_back(update_per(root[(int)root.size() - 1], 0, m - 1, 0, 0));
}
void find_ans(int v, int u, int lc, int pr, int tl, int tr, int k) {
  if (tl == tr) {
    if (all[v].sum + all[u].sum - all[lc].sum - all[pr].sum) ans.push_back(tl);
    return;
  }
  int now = all[all[v].l].sum + all[all[u].l].sum - all[all[lc].l].sum -
            all[all[pr].l].sum;
  int tm = (tl + tr) >> 1;
  if (now < k) {
    if (now) find_ans(all[v].l, all[u].l, all[lc].l, all[pr].l, tl, tm, now);
    if (k - now)
      find_ans(all[v].r, all[u].r, all[lc].r, all[pr].r, tm + 1, tr, k - now);
  } else {
    find_ans(all[v].l, all[u].l, all[lc].l, all[pr].l, tl, tm, k);
  }
}
void solve(int v, int u, int k) {
  int lc = lca(v, u), pr = up[lc][0];
  u = num[u], v = num[v], lc = num[lc], pr = num[pr];
  find_ans(v, u, lc, pr, 0, m - 1, k);
}
signed main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  int q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  g[0].push_back(1);
  g[1].push_back(0);
  for (int i = 0; i < m; i++) {
    int v;
    cin >> v;
    p[v].push_back(i);
  }
  add_node();
  root.push_back(0);
  build(0, 0, m - 1);
  dfs(0);
  for (int i = 0; i < q; i++) {
    int v, u, k;
    cin >> v >> u >> k;
    ans = {};
    solve(v, u, k);
    cout << (int)ans.size() << " ";
    for (auto j : ans) {
      cout << j + 1 << " ";
    }
    cout << '\n';
  }
}